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in this video we are going to see assignment problem by Hungarian method before we see the problem let us see what is assignment problem assignment problem is a special kind of transportation problem in which each source should have the capacity to fulfill the demand of any of the destination this is the basic concept of assignment problem so the main objective of assignment problem is to assign the jobs to the right operator such that the total processing time is minimized so that is the main objective of assignment problem there are two types of assignment problem first one is a balanced assignment problem another one is unbalanced assignment problem Saban's assignment problem means if the number of rows is equal to the number of column then it is called as balanced assignment problem otherwise an unbalanced assignment problem say in order to proceed the assignment problem first we need to check whether we have equal number of rows and equal number of columns then only we can proceed or otherwise we need to convert the unbalanced transportation problem into balanced transportation problem by adding either dummy row or dummy column then it will be converted as a balanced one then we can apply the following procedures say an assignment problem can be easily solved by applying Hungarian method so which consists of two phases the first phases row and column reduction this is the first phase after doing this step the second phases optimation of the problem these are the two important phases which comes under Hungarian method now I am going to explain the algorithm of Hungarian method along with one example problem look at this problem solve the following assignment problem using Hungarian method the matrix entries represent the processing times in Aza so there are five operators and five jobs in order to proceed the assignment problem we need to have equal number of rows and equal number of columns so look at this problem there are five rows and five columns are there so this one is balanced one if we have only four columns okay fibers then it is called as unbalanced it assignment problem in order to convert unbalanced into balance we need to add either dummy row or dummy column with zero values so this is the first procedure so in this problem we have five rows and five columns so this one is balanced one now we can go for Hungarian method look at the algorithm of Hungarian method the first phase is a row and column reduction okay so in this face there are two steps are there first one subtract the minimum value of each row from the entries of that row okay which means we need to find minimum value in each and every row separately in this case look at the first row nine eleven fourteen eleven seven so which one is the least value seven is the least value so we need to find row minimum so in the first two row minimum value is seven in the same way in the second row minimum value is six in the third row again six in the fourth row minimum value is nine here we have two nines you can take a nine and the fifth a row the least value is seven so we have got least value for each and every row the first step is subtract the minimum value of each and every row separately subtract the minimum value of each row from the entries of that row okay for that after finding the minimum value then you need to subtract this value from each and every element of that particular row so in the first row nine minus seven eleven minus seven fourteen minus seven eleven minus seven and 7 minus 7 will be getting these values two four seven four zero in the same way in the second row six minus six 15-6 13 minus 6 13 minus 6 10 minus 6 will be getting 0 9 7 7 4 in the same way find a new value for each and every row separately so this is the matrix a producting row minimum then go for second step step 2 is subtract the minimum value of each column from the entries of that particular column after applying the first step we got this matrix from this matrix we need to apply the second step that is called and reduction for that we need to find a minimum value of each and every column separately look at the first column 2 0 6 2 0 0 is the minimum value so put 0 in the second column also 0 is there so 0 is the minimum value in the third column also we have 0 so 0 is the minimum value in the fourth column 2 is a minimum value in the fifth column 0 is the minimum value so after finding call and minimization then apply the second step so what is the second step subtract the minimum value of each column from the entries of that particular column look at the first column the least value is 0 so 2 minus 0 0 - 0 6 - 0 2 - 0 0 - 0 then we will be getting new value the same thing only because yet 0 is there now so 2 - 0 - only 0 - 0 0 and a 6 - 0 6 and a 2 - 0 2 + 0 - 0 0 so if you get 0 value then the new value also reminds same look at the fourth column the least value is 2 ok 4 - 2 7 - 2 2 - 2 3 minus 2 and 3 minus 2 look at the new matrix at the first one - 5 0 1 and 1 ok in the last column leads to value 0 so 0 minus zie Oh 4-0 2-0 0-0 7-0 look at the new matrix zero four two zero seven so this is a new matrix after subtracting call and minimization so far we have done phase 1 that is row reduction and column reduction now let us see phase 2 okay the phase two is optimization of the problem in Phase two there are five steps or they step one and draw a minimum number of lines to cover all the zeros of the matrix so in order to apply the first step you have to follow two procedures the first procedure is row scanning and second procedure is column scanning for first one row scanning start from the first row ask the following question is there exactly one zero in that row if yes mark a square around that zero entry and draw a vertical line passing through that zero otherwise skip that particular row which means ensure only one zero in that particular row if you find more than one zero just keep that row go for the next room or otherwise if you find only one zero just put one square around that zero and draw a vertical line passing through that zero okay the same procedure can be applied for each and every row scanning and after scanning the last row check whether all the zeros are covered with the lines if yes go to step two otherwise do column scanning look at the procedure for column scanning the same thing start from the first column as the following question is there exactly one zero in that column if yes mark square around that zero entry and draw a horizontal line okay here you need to write here you need to draw a horizontal line passing through that zero otherwise skip that column which means ensure only one zero in that column okay if you find more than one zero means just keep that particular call go for the next two : if you find only one zero in that particular column ins then you can draw a square around the zero and then you can draw horizontal line passing through that zero and afterwards a scan the last column after scanning the last column check whether all the zeros or covered with lines okay so let me explain along with the problem after row reduction and call and reduction we have got the new matrix so in this matrix you need to draw a minimum number of lines to cover all the zeros of the matrix so we need to apply two procedures the first one is a row scanning and second one column scanning the first one is a row scanning okay so look at the first row just ensure that in the first row only one zero is there if you find more than one zero just skip that too then go for second row scanning okay if you find only one zero just put one square around that zero and then draw a vertical line passing through that zero the vertical line represent the entire column is deleted okay now go for second row look at the second row is there exactly one zero in that row yes so here we have only one zero just put one square around the zero and then draw a vertical line passing through this zero that is just delete this particular column by drawing vertical line okay so two rows are over now go for third row look at the third row just check whether we have only one zero but here we have two zero just keep go for the next step so now go to the fourth row in the fourth row we have only one 0 this 0 already deleted ok so just leave it here we have only one 0 and just put one square around the 0 and draw a vertical line passing through that zero ok so four rows are over noon go for the fifth row so in the fifth row no 0 okay so already one zero is there and that 0 is already deleted so we don't have any zeros that's all after scanning the last row check whether all the zeros are covered with lines if yes go to step 2 you need not go for column scanning but here still we have two zeros without covered with the lines so we need to go for column scanning according to the procedure row scanning is over now do the column scanning for the column scanning already two columns are over now start from the third column so look at the third column okay the third column just check whether we have got only one zero if we have more than one zero just keep that column then go for next column but here we have got only one zero in this particular column so put one square around the zero and saw a horizontal line for column scanning we need to draw a horizontal line for row scanning we need to draw vertical line now go for the fourth column look at the fourth column just check is there any one zero no already one zero is deleted so no zeros are there no so you cannot draw any line that's all so already first two column is deleted second column also deleted fifth the column also deleted we have got only two columns undeleted in that in the third column we have got one 0 and we have drawn horizontal line in the fourth column no 0 so all zeros are covered with minimum line so step one is over that is row scanning and column scanning is over look at step two check whether the number of square margarett is equal to the number of rows of the matrix if s go to step 5 that is optimality is reached otherwise go to next step that is go to third step let me explain with the problems after drawing a minimum line which covers all the zeros ensure that the number of square marker is equal to number of rows in this matrix say how many is are they 1 2 3 4 so how many number of squares for number of squares you must have equal number of rows so how many rows are there 1 2 3 4 5 rows so this is not equal okay if we get equal number of square an equal number of a row means then the optimality is reached it in this matrix the number of square is 4 & 5 Rosen this is not an equal 1 okay now we go for next step look at the third step identify the minimum value of the undeleted cell values after identifying the minimum value of the underrated cell value the first thing you need to add add the minimum undeleted cell value at the intersection points of the present matrix so first you need to add that value in the intersection point okay and second step is subtract the minimum undeleted cell value from all the undeleted cell values okay and third one is all other entries reminds same so these are the three things which comes under step 3 let me explain along with the problem now you need to form a new matrix by applying 3 steps the first thing is okay here we have undeleted cell values so among the undeleted cell value find which one is least value so 1 is the least value okay so we have got a minimum undeleted cell value okay after finding the minimum undeleted cell value just add that particular value in the intersection point so intersection point is this one is intersection point near the vertical line and horizontal line intersect each other so this point this intersection point and this point also intersection point and this point also intersection points are in the 3 intersection points add that least value so which one is least value 1 is the least value no so just take 1 and add this particular value in the intersection points the 6 will be converted into 7/7 will be converted into a to will be converted into three okay so we need to add this value at the intersection points that is the first thing and second step is a-- okay just take the minimum undeleted value and subtract this value from all other undeleted values here we have seven two seven five one one seven one these are the undeleted values from each and every undeleted values we need to subtract this minimum value that is a step two and third one is for all other values remind same that is 2 4 0 9 2 0 0 5 7 0 4 0 these are the other values reminds same just find the least value of the undeleted cell that is 1 is the least value no take 1 and apply the first step first step is add this value at the intersection pointer so here 6 will become 7 7 will become 7 plus 1 H 2 will become 3 2 plus 1 3 ok and then subtract this one from all other undeleted values of 7 minus 1 6 2 minus 1 1 7 minus 1 6 5 minus 1 4 1 minus 1 0 1 minus 1 0 7 minus 1 6 1 minus 1 0 ok we have added 1 at the intersection point and we have subtracted at the undeleted values now just copy all other values as it is two zero two zero two zero two zero four nine zero five four nine zero five then zero four zero seven zero four zero seven here zero zero zero zero okay so this is pneumatics after applying three procedures which comes under step three we have done step three with three procedures now go for step four Step four is now you have to start from step one already we know very well what is a step one row scanning and column scanning so we needed to a minimum line which covers all the zeros for that you have to apply row scanning and column scanning so now I'm going to do row scanning look at the first row and sure that is only one zero in that particular row yes we have only one zero so draw a vertical line and go for the second row in the second row also we have got only one zero so put one square around it and pass and draw a vertical line passing through that zero and go for third row the third row we have got two zeros just skip this row and go for fourth one in the fourth row again we have three zeros just skip and go for the fifth row in the fifth row already one zero is jelly deck we have only one zero just put one square and draw a vertical line so row scannings are over just check whether all the zeros or delete that no still we have three zeros now go for column scanning so the first column already deleted no go for second column so the second column ensure only one zero yes we have only one zero put one square around it and draw a horizontal line okay horizontal line drawn horizontal line okay then go for third column in the third column this zero is already deleted so now we have only one zero so put one square around it and draw horizontal line and already fourth column is the column delete that you need not go and scan anything now just check whether all the zeros or deleted okay and all the zeros are deleted now apply the rule what is a rule for optimality just check the number of square is quite a number of row so how many squares are there 1 2 3 4 & 5 the number of square mark is 5 which is equal to number of rows of the matrix hence the solution is optimal and feasible now write down the optimal solution for job 1 2 3 4 5 we need to write the operator look at the first one for first job operator is 5 for second job operator is 1 for job 3 operator is 3 for job for operator is 2 for job fine operator is 4 now right this time this you need to write from the problem look at the problem for us 2 1 5 9 5 for job 1 operator is 5 for job - operator is 1 for job 3 operator is 3 for job for operator is 2 for job Phi operator is 4 4 2 1 4 3 3 4 4 2 4 5 4 now you have to write the timing for 1 this is the time for each and every job okay for the first one 7 for the second one 6 for the third one 6 for the fourth one 9 for the 5110 7 6 6 9 10 total the total processing time is 38 as this is the way to solve assignment problem hope you all understood this concept thank you